What are the solutions to the equation 3x^2+2x+8=0 ?

The solutions to the equation 3x^2+2x+8=0 are x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3

Find the zeros of 3x^2+2x+8=0

The zeros of 3x^2+2x+8=0 are x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3

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3 x^{2} + 2 x + 8 = 0

x = - \frac{1}{3} + i \frac{23}{3}, x = - \frac{1}{3} - i \frac{23}{3}

Solution steps

3 x^{2} + 2 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

2^{2} - 4 \cdot 3 \cdot 8 : 223 i

x_{1, 2} = \frac{- 2 \pm 2 23 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 23 i}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 23 i}{2 \cdot 3}

x = \frac{- 2 + 2 23 i}{2 \cdot 3} : - \frac{1}{3} + i \frac{23}{3}

x = \frac{- 2 - 2 23 i}{2 \cdot 3} : - \frac{1}{3} - i \frac{23}{3}

The solutions to the quadratic equation are:

x = - \frac{1}{3} + i \frac{23}{3}, x = - \frac{1}{3} - i \frac{23}{3}

(4 x + 2)^{2}

31

8 c - 6 \leq - 5 c - 3

x^{2} y^{2} - 64

\frac{( 3 r ^{- 2} s ^{3} t ^{0} ) ^{- 3}}{3 rs}

What are the solutions to the equation 3x^2+2x+8=0 ?
The solutions to the equation 3x^2+2x+8=0 are x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3
Find the zeros of 3x^2+2x+8=0
The zeros of 3x^2+2x+8=0 are x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3